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Math Help - how would I apply Bayes theorem?

  1. #1
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    how would I apply Bayes theorem?

    Machines I and II respectively produce 70% and 30% of a factory’s output. Machine I produces 2% defectives and machine II produces 3% defectives. If an item is not defective, what is the chance it was produced by machine I?

    Which event would be A and which one B?
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  2. #2
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    what about this one matheagle? i thanked you for your other ones
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  3. #3
    MHF Contributor matheagle's Avatar
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    I will let A stand for machine one and B for machine 2.

    You have P(A)=.7 and P(B)=.3 as the partition.

    Next you have P(D|A)=.02 and P(D|B)=.03 where D stands for defective.

    So the probability of defective is

    P(D)=P(DA)+P(DB)= P(D|A)P(A)+P(D|B)P(B)

    =(.02)(.7)+(.03)(.3)=.014+.009=.023

    Hence the probability of a NONdefective is .977.

    Let N stand for nondefective. The question here is P(A|N)={P(AN)\over P(N)}.

    We have P(N)=.977 and from .7=P(A)=P(AN)+P(AD)

    we can obtain P(AN) since P(AD)=(.02)(.7) from above.

    I get .702149437 for the final answer, which seems reasonable since P(A)=.7 isn't far away.
    Last edited by matheagle; May 26th 2009 at 10:35 PM.
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  4. #4
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    is the answer 68.6% and if not can you expain the answer to me

    and then what is the chance that is was produced by machine I
    was my answer correct?
    Last edited by mr fantastic; May 26th 2009 at 02:23 AM. Reason: Merged posts
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  5. #5
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    Quote Originally Posted by crafty View Post
    is the answer 68.6% and if not can you expain the answer to me

    and then what is the chance that is was produced by machine I
    was my answer correct?
    You have been given all the details of the calculation and you have been given the answer (which is NOT equivalent to 68.6% by the way). I suggest that you carefully review the reply you got and then go back and thoroughly review this material.
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